Impedance

There's a good chance you've heard of Ohm's Law - Voltage equals Current multiplied by Resistance.

$$V=I \times R$$

We can think of a simple battery driven Direct Current circuit, a resistor and LED for example, where we are pushing a current across and limited by a resistor. For a given voltage, the larger the resistor, the less current will pass through.

It's also likely you've heard of impedance, and things like that "Guitars are high impedance, or high-Z" - but what does that mean? Well, impedance is also a resistance, and even shares the same Unit of $\Omega$ (Ohms). In audio, our signal is an Alternating Current - impedance is the opposition to the flow of current in a circuit to an AC signal, and determines how efficiently power moves between gear. It can also be frequency dependent!

Impedance is usually given the symbol $Z$. There is also a component of interaction from capacitors and inductors call reactance ($X$), and for simplicity we might say "impedance is the resistance to an AC signal", but it is technically a complex number which is the sum of the reactance and standard resistance.

$$Z = R \times \imath X$$

It's All Connected

When we say "resistance of a circuit", we mean the total circuit - so this can also be viewed as the "black box of our gear", e.g. a guitar amp is a "circuit" and so is the speaker cab plugged into it. Gear that generates a voltage has an output impedance, and gear that accepts a voltage has an input impedance.

Let's take a moment to think about a mis-matched guitar amp and cab example (a power transfer issue). Have you ever had something like an amp with a $4 \Omega$ output impedance, and were confused about which cab was safe to use - a lower $2 \Omega$ one, or a higher $8 \Omega$ one? If we think about impedance as "resistance to our AC signal", and "more resistance means less power transfer", then we can reason that the amp will have a harder time pushing the signal through the larger one - which is what we want! It will be too easy to push a signal through the $2 \Omega$ one, which would result it too much current/over heating/damaging our gear. It's safer to push a signal into a higher impedance.

Now let's consider a microphone that's sitting in front of that cab. We don't really want to transfer a lot of power from our mic to our preamp (i.e. current), but we do want a strong clean voltage signal. This is now a voltage bridging issue. We can do this by matching a low impedance mic with a much higher impedance preamp - this has lead to a typical rule of about 10 to 1 higher - so a $100 \Omega$ mic would work well with a $10 k\Omega$ preamp.

But what about those high-Z guitar pickups going into the amp? This is still a voltage bridging issue. A guitar amp is expecting those high impedance picks of 20-50 $k\Omega$, but what happens when you plug into that $10 k\Omega$ interface you're using with the mic? It's lower! Welcome to "tone suck"! This is why lot's of interfaces have a setting (or dedicated input) for high-Z gear, which has an input impedance in that 10:1 ratio - typically in the $M\Omega$ range.

Wrapping Up

Knowing the output and input impedance of your gear is important to get the best signal - and so is knowing if you're concerned with power transfer or voltage bridging. An end-to-end guitar rig is a good example to keep in mind for this.